I need to know if an expression contain a radical in it. For example, expr:=1/x*sqrt(x*y)  or expr:=3*(x*y)^(2/3)+5; or expr := sqrt(5), etc.. all these will return true, but for expr:=3+sin(x) it will return false.

I need to find if the expression has radical, becuase then I want to do special simplify if that is the case.

I do not want to use pattern matching, since the expression can be complicated, and thought to ask if there might be a simpler way, one of those hidden Maple commands which can do it.

I looked at  "has", "hasfun". with(PolynomialIdeals) has function IsRadical but it is only for polynomials.

Is there a way other than using patmatch?  

is possible to solve these partial differential equations in maple via pdsolve?

Thanks

Hello,

I tried to operate a transformation of some random variables but Maple doesn't compute the PDF of the outcome RV. Do you have a suggestion on how I should edit my code ? (the goal is to get the PDF of S)

Thanks in advance,

Antoine

Way i get this error

plot([x, cos(x), x = -Pi .. Pi, y = -.5 .. .5])

Error, (in plot) incorrect first argument [x, cos(x), x = -Pi .. Pi, y = -.5 .. .5]
 

In an unrelated thread, I provided the OP with some 1-D code, which contained the Array definition

TC:= Array(0...1001, fill=0)

Note the existence of three '.' characters in the range specification. This was a typo on my part, or my '.' key bounced, or something. The code containing the above definition "worked" with no problem, which, presumably, was why I didn't notice.

The Maple help does state (my emphasis)

Note that more than two dots in succession are also parsed as the range (..) operator.

although I wasn't making use of this fact - I just screwed up when typing the original.

The OP preferred to use 2-D input, and used cut-and-paste to transfer the above code, resulting in 2-D input, which is where the fun started. It seems(?) that when using 2-D input, more than two dots in succession is only interpreted as a straightforward range, if the total number of dots is even.

If the total number of dots is odd, then it appears(?) as if the 'final' dot is associated with the second number in the range as a 'decimal point', (so producing .1001 in the above example). This is then 'coerced/rounded' to an integer - ie it becomes '0', and the above Array definition is interpreted as

TC:= Array(0..0, fill=0)

Consequences in the following code are left to your imagination

Worth an SCR?

Dear Users!

Hope you would be fine with everthing. I am going to draw a closed figure in maple for this I defined 13 function and then plot them combine. But function 13 "F13" not plotted as I required. I need it plot vertically but it plot horizentaly. Please see the attachment and try to fix my problem. I am waiting your response. Thanks in advance.

Functions.mw

how to adjust the height and width of plot in maple 13

plot( x[3]^5, caption = typeset("\n A plot of %1.", x[3]^5), captionfont=[times, 20] );

hellow,

can any body help me to  increase the size of the caption in maple plot

plot( x[3]^5, caption = typeset("A plot of %1.", x[3]^5) );

Can anyone  produce these diagram?  Please Read the theory in:

   https://en.wikipedia.org/wiki/Logistic_map.

Wikipedia pages that explain bifurcation diagrams and attractors in more elementary contexts.

See the bifurcation diagram in the picture

   https://en.wikipedia.org/wiki/Logistic_map#/media/File:Logistic_Bifurcation_map_High_Resolution.png

 

Download aquestion.mw

Hi. How can I see the code of a graph already plotted.

Thanks

hi
can somebody hep me
i dont know how to add a neww member to an existing set!?
i made an empty set and in my loop for each prime answer i have to add a new memeber to my set but i dont know how!

Is this a correct way to define a piecewise function in which substitution occurs, or are there better ones?

Data := [L=1000, F=1000, E=206000, d0=10];

Diameters := X -> eval(subs(Data, piecewise(x<L/2, d0 ,x<2/3*L, 2*d0 ,x>2/3*L, 1.5*d0)), x=X); 
 

Hello.

Regarding to my previous question I'd like to speed up calculations of the expression. 

restart;
tt := -0.689609e-3; T_c := .242731; mu := .365908; k := 1;
R1 := a*tanh((a^2-mu)/(2*T_c))*ln((2*a^2+2*a*q+q^2-2*mu-(I*2)*Pi*N)/(2*a^2-2*a*q+q^2-2*mu-(I*2)*Pi*N))/q-2;
R2 := Int(R1, a = 0 .. 10000);
R3 := q*ln((-q^2-k^2+mu+I*(2*N*Pi*T_c-(2*m+1)*Pi*T_c)+k*q)/(-q^2-k^2+mu+I*(2*N*Pi*T_c-(2*m+1)*Pi*T_c)-k*q))/(k*(tt+R2));
R4 := Sum(R3, N = -100 .. 100);
m := 1;
R5 := Int(R4, q = 0.1e-2 .. 10000);
R6 := evalf(R5);

Here I have integration procedure inside the expression R3, then the summation over the integer parameter N and then finally the integration again.

Is it possible to speed up calculations of this cumbersome expression? Or actually was I correct to write this simple code?

Thank you in a advance.

K := simplify(C, 'size');
      5                                                  4   
lambda  + (-a__11 - a__33 - a__44 - a__55 - a__22) lambda  + 

  ((a__44 + a__33 + a__22 + a__11) a__55

   + a__44 (a__33 + a__22 + a__11) + (a__33 + a__22) a__11

                                      3                    
   + a__33 a__22 - a__32 a__23) lambda  + (((-a__33 - a__22

   - a__11) a__44 + (-a__33 - a__22) a__11 - a__33 a__22

   + a__32 a__23) a__55

   + ((-a__33 - a__22) a__11 - a__33 a__22 + a__32 a__23) a__44

   + (-a__22 a__33 + a__23 a__32) a__11

                                              2            
   - a__32 (a__13 a__21 + a__24 a__43)) lambda  + ((((a__33

   + a__22) a__11 + a__33 a__22 - a__32 a__23) a__44

   + (a__22 a__33 - a__23 a__32) a__11

   + a__32 (a__13 a__21 + a__24 a__43)) a__55

   + ((a__22 a__33 - a__23 a__32) a__11 + a__13 a__21 a__32) a__44

   + a__32 (a__11 a__43 a__24 - a__21 (a__14 a__43 + a__15 a__53)

  )) lambda + (((-a__22 a__33 + a__23 a__32) a__11

   - a__13 a__21 a__32) a__44

   - a__43 a__32 (a__11 a__24 - a__14 a__21)) a__55

   - a__15 a__21 a__32 (a__43 a__54 - a__44 a__53)

u := [coeffs(K, [lambda], 'l')];
[(((-a__22 a__33 + a__23 a__32) a__11 - a__13 a__21 a__32) a__44

   - a__43 a__32 (a__11 a__24 - a__14 a__21)) a__55

   - a__15 a__21 a__32 (a__43 a__54 - a__44 a__53), 1, 

  -a__11 - a__33 - a__44 - a__55 - a__22, (a__44 + a__33 + a__22

   + a__11) a__55 + a__44 (a__33 + a__22 + a__11)

   + (a__33 + a__22) a__11 + a__33 a__22 - a__32 a__23, ((-a__33

   - a__22 - a__11) a__44 + (-a__33 - a__22) a__11 - a__33 a__22

   + a__32 a__23) a__55

   + ((-a__33 - a__22) a__11 - a__33 a__22 + a__32 a__23) a__44

   + (-a__22 a__33 + a__23 a__32) a__11

   - a__32 (a__13 a__21 + a__24 a__43), (((a__33 + a__22) a__11

   + a__33 a__22 - a__32 a__23) a__44

   + (a__22 a__33 - a__23 a__32) a__11

   + a__32 (a__13 a__21 + a__24 a__43)) a__55

   + ((a__22 a__33 - a__23 a__32) a__11 + a__13 a__21 a__32) a__44

   + a__32 (a__11 a__43 a__24 - a__21 (a__14 a__43 + a__15 a__53)

  )]
u[1] = C__5;
(((-a__22 a__33 + a__23 a__32) a__11 - a__13 a__21 a__32) a__44

   - a__43 a__32 (a__11 a__24 - a__14 a__21)) a__55

   - a__15 a__21 a__32 (a__43 a__54 - a__44 a__53) = C__5
C__1 = u[3];
         C__1 = -a__11 - a__33 - a__44 - a__55 - a__22
C__2 = u[4];
   C__2 = (a__44 + a__33 + a__22 + a__11) a__55

      + a__44 (a__33 + a__22 + a__11) + (a__33 + a__22) a__11

      + a__33 a__22 - a__32 a__23
C__3 = u[5];
C__3 = ((-a__33 - a__22 - a__11) a__44 + (-a__33 - a__22) a__11

   - a__33 a__22 + a__32 a__23) a__55

   + ((-a__33 - a__22) a__11 - a__33 a__22 + a__32 a__23) a__44

   + (-a__22 a__33 + a__23 a__32) a__11

   - a__32 (a__13 a__21 + a__24 a__43)
C__4 = u[6];
C__4 = (((a__33 + a__22) a__11 + a__33 a__22 - a__32 a__23) a__44

   + (a__22 a__33 - a__23 a__32) a__11

   + a__32 (a__13 a__21 + a__24 a__43)) a__55

   + ((a__22 a__33 - a__23 a__32) a__11 + a__13 a__21 a__32) a__44

   + a__32 (a__11 a__43 a__24 - a__21 (a__14 a__43 + a__15 a__53)

  )